All categories

2 years ago

Please help me to find area of semicircle

Mathematics

1 answer:

mojhsa [17]2 years ago

5 0

Answer:

Use a calculator and use the formula 1/2 × pi × (radius squared)

You might be interested in

In an Algebra II class with 135 students, the final exam scores have a mean of 72.3 and standard deviation 6.5. The exams on the

gregori [183]

Population = 135 students
Mean score = 72.3
Standard deviation of the scores = 6.5

Part (a): Students from 2SD and 3SD above the mean
2SD below and above the mean includes 95% of the population while 3SD includes 99.7% of the population.
95% of population = 0.95*135 ≈ 129 students
99.7% of population = 0.997*135 ≈ 135 students

Therefore, number of students from 2SD to 3SD above and below the bean = 135 - 129 = 6 students.
In this regard, Students between 2SD and 3SD above the mean = 6/2 = 3 students

Part (b): Students who scored between 65.8 and 72.3
The first step is to calculate Z values
That is,
Z = (mean-X)/SD
Z at 65.8 = (72.3-65.8)/6.5 = 1
Z at 72.3 = (72.3-72.3)/6.5 = 0
Second step is to find the percentages at the Z values from Z table.
That is,
Percentage of population at Z(65.8) = 0.8413 = 84.13%
Percentage of population at (Z(72.3) = 0.5 = 50%
Third step is to calculate number of students at each percentage.
That is,
At 84.13%, number of students = 0.8413*135 ≈ 114
At 50%, number of students = 0.5*135 ≈ 68

Therefore, students who scored between 65.8 and 72.3 = 114-68 = 46 students

4 0

2 years ago

Jasmine created this spinner to predict whether a randomly selected student in her school will be traveling during the upcoming

riadik2000 [5.3K]

Answer: 9

Step-by-step explanation:

There are 8 equal sections which are meant to reflect the data Jasmine collected from surveying the students.

This means that each section represents;

= 36/8

= 4.5 students.

There are 2 sections that represent the number of students who said they would be traveling during the holiday break.

= 4.5 * 2

= 9 students

4 0

2 years ago

Read 2 more answers

Solve and state any extraneous solutions

Mice21 [21]

$\frac{3x}{x - 1} + 2 = \frac{3}{x - 1}$

$\frac{3x}{x - 1} + \frac{2(x - 1) }{ x - 1} = \frac{3}{x - 1}$

$\frac{3x}{x - 1} + \frac{2x - 2}{ x - 1} = \frac{3}{x - 1}$

$\frac{3x + 2x - 2}{ x - 1} = \frac{3}{x - 1}$

$3x + 2x - 2 = 3$

$5x - 2 = 3$

$5x = 3 + 2$

$5x = 5$

$x = \frac{5}{5}$

$x = 1$

3 0

2 years ago

Which of the following rational functions is graphed below?

jolli1 [7]

Because we have a vertical asymptote at x = -1, we conclude that the correct option is C.. Which of the following rational functions is graphed below? In the graph, we can see that we have a vertical asymptote at x = -1.. …

8 0

1 month ago

Find the volume and area for the objects shown and answer Question

klio [65]

Step-by-step explanation:

You must write formulas regarding the volume and surface area of the given solids.

$\bold{\#1\ Rectangular\ prism:}\\\\V=lwh\\SA=2lw+2lh+2wh=2(lw+lh+wh)\\\\\bold{\#2\ Cylinder:}\\\\V=\pi r^2h\\SA=2\pi r^2+2\pi rh=2\pir(r+h)\\\\\bold{\#3\ Sphere:}\\\\V=\dfrac{4}{3}\pi r^3\\SA=4\pi r^2$

$\bold{\#4\ Cone:}\\\\V=\dfrac{1}{3}\pi r^2h\\\\\text{we need calculate the length of a slant length}\ l\\\text{use the Pythagorean theorem:}\\\\l^2=r^2+h^2\to l=\sqrt{r^2+h^2}\\\\SA=\pi r^2+\pi rl=\pi r^2+\pi r\sqrt{r^2+h^2}=\pi r(r+\sqrt{r^2+h^2})\\\\\bold{\#5\ Rectangular\ Pyramid:}\\\\V=\dfrac{1}{3}lwh\\\\$

$\\\text{we need to calculate the height of two different side walls}\ h_1\ \text{and}\ h_2\\\text{use the Pythagorean theorem:}\\\\h_1^2=\left(\dfrac{l}{2}\right)^2+h^2\to h_1=\sqrt{\left(\dfrac{l}{2}\right)^2+h^2}=\sqrt{\dfrac{l^2}{4}+h^2}=\sqrt{\dfrac{l^2}{4}+\dfrac{4h^2}{4}}\\\\h_1=\sqrt{\dfrac{l^2+4h^2}{4}}=\dfrac{\sqrt{l^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{l^2+4h^2}}{2}$

$\\\\h_2^2=\left(\dfrac{w}{2}\right)^2+h^2\to h_2=\sqrt{\left(\dfrac{w}{2}\right)^2+h^2}=\sqrt{\dfrac{w^2}{4}+h^2}=\sqrt{\dfrac{w^2}{4}+\dfrac{4h^2}{4}}\\\\h_2=\sqrt{\dfrac{w^2+4h^2}{4}}=\dfrac{\sqrt{w^2+4h^2}}{\sqrt4}=\dfrac{\sqrt{w^2+4h^2}}{2}$

$SA=lw+2\cdot\dfrac{lh_1}{2}+2\cdot\dfrac{wh_2}{2}\\\\SA=lw+2\!\!\!\!\diagup\cdot\dfrac{l\cdot\frac{\sqrt{l^2+4h^2}}{2}}{2\!\!\!\!\diagup}+2\!\!\!\!\diagup\cdot\dfrac{w\cdot\frac{\sqrt{w^2+4h^2}}{2}}{2\!\!\!\!\diagup}\\\\SA=lw+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw}{2}+\dfrac{l\sqrt{l^2+4h^2}}{2}+\dfrac{w\sqrt{w^2+4h^2}}{2}\\\\SA=\dfrac{2lw+l\sqrt{l^2+4h^2}+w\sqrt{w^2+4h^2}}{2}$

6 0

2 years ago